Statics: particle mechanics versus rigid body mechanics

1. In particle mechanics where the notion of rotation does not apply, particles are said to be in static equilibrium when the sum of the external forces acting on the particle of interest in all directions equals zero.

2. In rigid body mechanics where the notion of rotation applies (or at least the concept of invariant distances between points), rigid bodies are said to be in static equilibrium when (1) the sum of the external forces in all directions equals zero, and (2) the sum of the external moments in all directions equals zero.

My question is: in rigid body mechanics, where does the equation on moments come from? Is it a principle similar to Newton's law in particle mechanics or can it be proven from other assumptions (like, for instance, the invariance of distances separating two points)

• You have two options depending on how you model the rigid body. If you model it as an assemblage of point masses, then yes you can derive these rules from Newton's laws. Or, if you want to distinguish between rigid continua from rigidly connected points, you can take Euler's laws of motion (which extend Newton's laws to continua) as axioms. – David H Aug 26 '13 at 21:04
• I'm afraid I don't. But it is a standard theorem that is usually proved in intermediate classical mechanics classes and textbooks. See chapter 1 of Goldstein's Classical Mechanics for example. – David H Aug 26 '13 at 22:20

One way to see a rigid body is a collection of infinity particles such that they preserve the distances between themselves (can only translate and rotate). So, we have a constraint, and then internal forces that maintain this constraint.

This internal forces are dependent on the forces that you apply in the system. If you apply a force in one point of the object (one particle), then all other points would feel a force such that the accelerations of the hole system don't destroy the constraint. Rigid bodies need to have energy stored there to do that.

If you have a static homogeneous rigid body with the center of mass $\vec{r}_{cm}=\vec{0}$, applying a force $\vec{f}(\vec{r}_0)$ in some point $\vec{r}_0$ such that the force is paralel to $\vec{r}_0$, then all the points of the body need to feel this same force to preserve the constraint. The result of this force is an aceleration of the whole system generating a translation.

If the force $\vec{f}(\vec{r}_0)$ is not parallel to $\vec{r}_0$, then all the points need to feel a force $\vec{f}(\vec{r})$ such that

$$\vec{r}\times\vec{f}(\vec{r})=\vec{r}_0\times\vec{f}(\vec{r}_0)$$.

because the constraint is mathematically translated to $d\vec{r}=d\vec{\theta}_0\times\vec{r}$ for every $\vec{r}$ and the same $d\vec{\theta}_0$ (see here for understand why).

With this we can define a quantity called Torque associated for a force $f$ applied to a point $\vec{r}$ of the rigid body by:

$$\vec{\tau}=\vec{r}\times\vec{f}(\vec{r})$$

So, if we have a force $\vec{f}_1$ applied at $\vec{r}_1$ and $\vec{f}_2$ applied at $\vec{r}_2$ we can see that the net force at point $\vec{r}$ need to obey:

$$\vec{f}(\vec{r})\times\vec{r}=\vec{f}_1\times\vec{r}_1+\vec{f}_2\times\vec{r}_2$$

Then we may define a net torque:

$$\vec{\tau}_{net}=\sum_{j}\vec{f}_{j}\times\vec{r}_{j}$$

And, if $\vec{\tau}_{net}=\vec{0}$, then the body does not rotate because at each point $\vec{f}(\vec{r})\times\vec{r}=0$ and for $\vec{r}\neq \vec{0}=\vec{r}_{cm}$ we conclude that the force $\vec{f}(\vec{r})$ is paralell to $\vec{r}$.